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so the question asks to find a sufficient statistic based on a sample of size n, where $X \sim f_{\theta}(x)$ and \begin{equation} f_{\theta}(x) = P_{\theta}(X=x) = c(\theta) 2^{-x/\theta}, \quad x = \theta, \theta+1,..., \theta >0 \end{equation} and \begin{equation} c(\theta) = 2^{1-1/\theta}(2^{1/\theta}-1) \end{equation} I derived that the joint pmf is \begin{equation} P_{\theta}(X_1 = x_1,...,X_n = x_n) = c(\theta)^{n}2^{-1/\theta \sum x_{i}}\mathbb{I}_{\{X_{(1)}\geqslant \theta\}} \end{equation} My question is, in this case, is a sufficient statistic $T(x) = (\sum x_{i}, X_{(1)})$ or both $\sum x_{i}$ and $X_{(1)}$ can serve as a sufficient statistic? Thanks so much for your time and consideration.

Andrew Liu
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    Yes, $T(X_1,\ldots,X_n)=(\sum X_i, X_{(1)})$ is sufficient for $\theta$ but that does not necessarily imply that the components of $T$ are themselves sufficient. – StubbornAtom Aug 30 '18 at 18:00

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You write that the question asks to find “the sufficient statistic”, but there is never only one sufficient statistic – in particular, any non-zero multiple of a sufficient statistic is also a sufficient statistic. The question probably meant to ask for “a sufficient statistic”.

The statistic you found is indeed sufficient. Neither of its components is sufficient. By definition, if a statistic is sufficient, the likelihood of the data depends on the data only through the statistic. By contrast, different data can lead to different $\sum_iX_i$ (and thus a different likelihood) but the same $X_{(1)}$, and vice versa. Thus neither of these components is sufficient by itself.

joriki
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  • You are right. The question asks to find a sufficient statistic. I should edit it. And thanks for the clarification. – Andrew Liu Aug 30 '18 at 21:04